Question: $ f(x)=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+3}}}{\left( 2n+1 \right)!}}$ Find the power series of $f''(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+2 \right)(2n+1){{x}^{2n+1}}}{\left( 2n +1\right)!}}$ (Choice B) B $\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n+1}}\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n +1\right)!}}$ (Choice C) C $\sum\limits_{n=0}^{\infty }{\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n +1\right)!}}$ (Choice D) D $\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n +1\right)!}}$ (Choice E) E $\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n -1\right)!}}$
Explanation: $\begin{aligned} &\phantom{=} f\left( x \right) \\\\ &=x^3-\dfrac{{{x}^{5}}}{3!}+...+{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+3}}}{\left( 2n +1\right)!}+... \\\\ &=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+3}}}{\left( 2n+1 \right)!}} \\\\ \\\\ &\phantom{=} {f}\ '\left( x \right) \\\\ &=3x^2-\dfrac{5{{x}^{4}}}{3!}+...+{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right){{x}^{2n+2}}}{\left( 2n +1\right)!}+... \\\\ &=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right){{x}^{2n+2}}}{\left( 2n +1\right)!}} \end{aligned}$ $\begin{aligned} &\phantom{=} {f}\ ''\left( x \right) \\\\ &=6x-\dfrac{20{{x}^{3}}}{3!}+...+{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n +1\right)!}+... \\\\ &=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+3 \right)(2n+2){{x}^{2n+1}}}{\left( 2n +1\right)!}} \end{aligned}$